1. 8- 1
Chapter 8
The Theory of NP-
Completeness

2. 8- 2
 P: the class of problems which can be solved
by a deterministic polynomial algorithm.
 NP : the class of decision problem which can
be solved by a non-deterministic polynomial
algorithm.
 NP-hard: the class of problems to which every
NP problem reduces.
 NP-complete (NPC): the class of problems
which are NP-hard and belong to NP.

3. 8- 3
Some concepts of NPC
 Definition of reduction: Problem A reduces to
problem B (A ∝ B) iff A can be solved by a
deterministic polynomial time algorithm using
a deterministic algorithm that solves B in
polynomial time.
 Up to now, none of the NPC problems can be
solved by a deterministic polynomial time
algorithm in the worst case.
 It does not seem to have any polynomial time
algorithm to solve the NPC problems.

4. 8- 4
 The theory of NP-completeness always
considers the worst case.
 The lower bound of any NPC problem seems
to be in the order of an exponential function.
 Not all NP problems are difficult. (e.g. the
MST problem is an NP problem.)
 If A, B ∈ NPC, then A ∝ B and B ∝ A.
 Theory of NP-completeness:
If any NPC problem can be solved in polynomial
time, then all NP problems can be solved in
polynomial time. (NP = P)

5. 8- 5
Decision problems
 The solution is simply “Yes” or “No”.
 Optimization problems are more difficult.
 e.g. the traveling salesperson problem
 Optimization version:
Find the shortest tour
 Decision version:
Is there a tour whose total length is less than or
equal to a constant c ?

6. 8- 6
Solving an optimization problem by a
decision algorithm :
 Solving TSP optimization
problem by a decision
algorithm :
 Give c1
and test (decision algorithm)
Give c2
and test (decision algorithm)

Give cn
and test (decision algorithm)
 We can easily find the smallest ci

7. 8- 7
The satisfiability problem
 The satisfiability problem
 The logical formula :
x1
v x2
v x3
& - x1
& - x2
the assignment :
x1
← F , x2
← F , x3
← T
will make the above formula true .
(-x1
, -x2
, x3
) represents x1
← F , x2
← F , x3
← T

8. 8- 8
 If there is at least one assignment which
satisfies a formula, then we say that this
formula is satisfiable; otherwise, it is
unsatisfiable.
 An unsatisfiable formula :
x1
v x2
& x1
v -x2
& -x1
v x2
& -x1
v -x2

9. 8- 9
 Definition of the satisfiability problem: Given
a Boolean formula, determine whether this
formula is satisfiable or not.
 A literal : xi
or -xi
 A clause : x1
v x2
v -x3
≡ Ci
 A formula : conjunctive normal form (CNF)
C1
& C2
& … & Cm

10. 8- 10
 Resolution principle
C1
: x1
v x2
C2
: -x1
v x3
⇒ C3
: x2
v x3
 From C1
&C2
, we can
obtain C3
, and C3
can
be added into the
formula.
 The formula becomes:
C1
&C2
&C3
The resolution principle
x1
x2
x3 C1
&C2
C3
0 0 0 0 0
0 0 1 0 1
0 1 0 1 1
0 1 1 1 1
1 0 0 0 0
1 0 1 1 1
1 1 0 0 1
1 1 1 1 1

11. 8- 11
 Another example of resolution principle
C1
: -x1
v -x2
v x3
C2
: x1
v x4
⇒ C3
: -x2
v x3
v x4
 If no new clauses can be deduced, then
it is satisfiable.
-x1
v -x2
v x3
(1)
x1
(2)
x2
(3)
(1) & (2) -x2
v x3
(4)
(4) & (3) x3
(5)
(1) & (3) -x1
v x3
(6)

12. 8- 12
 If an empty clause is deduced, then it is
unsatisfiable.
- x1
v -x2
v x3
(1)
x1
v -x2
(2)
x2
(3)
- x3
(4)
⇓ deduce
(1) & (2) -x2
v x3
(5)
(4) & (5) -x2
(6)
(6) & (3) □ (7)

13. 8- 13
Semantic tree
 In a semantic tree, each
path from the root to a
leaf node represents a
class of assignments.
 If each leaf node is
attached with a clause,
then it is unsatisfiable.

14. 8- 14
Nondeterministic algorithms
 A nondeterminstic algorithm consists of
phase 1: guessing
phase 2: checking
 If the checking stage of a nondeterministic
algorithm is of polynomial time-complexity, then
this algorithm is called an NP (nondeterministic
polynomial) algorithm.
 NP problems : (must be decision problems)
 e.g. searching, MST
sorting
satisfiability problem (SAT)
traveling salesperson problem (TSP)

15. 8- 15
Decision problems
 Decision version of sorting:
Given a1
, a2
,…, an
and c, is there a permutation
of ai
′
s ( a1
′
, a2
′
, … ,an
′
) such that∣a2
′
–a1
′
∣+∣a3
′
–a2
′
∣+
… +∣an
′
–an-1
′
∣ ＜ c ?
 Not all decision problems are NP problems
 E.g. halting problem :

Given a program with a certain input data, will
the program terminate or not?

NP-hard

Undecidable

16. 8- 16
Nondeterministic operations
and functions
[Horowitz 1998]
 Choice(S) : arbitrarily chooses one of the elements in set
S
 Failure : an unsuccessful completion
 Success : a successful completion
 Nonderministic searching algorithm:
j ← choice(1 : n) /* guessing */
if A(j) = x then success /* checking */
else failure

17. 8- 17
 A nondeterministic algorithm terminates
unsuccessfully iff there does not exist a set of
choices leading to a success signal.
 The time required for choice(1 : n) is O(1).
 A deterministic interpretation of a non-
deterministic algorithm can be made by
allowing unbounded parallelism in
computation.

18. 8- 18
Nondeterministic sorting
B ← 0
/* guessing */
for i = 1 to n do
j ← choice(1 : n)
if B[j] ≠ 0 then failure
B[j] = A[i]
/* checking */
for i = 1 to n-1 do
if B[i] > B[i+1] then failure
success

19. 8- 19
Nondeterministic SAT
/* guessing */
for i = 1 to n do
xi ← choice( true, false )
/* checking */
if E(x1, x2, … ,xn) is true then success
else failure

20. 8- 20
Cook’s theorem
NP = P iff the satisfiability
problem is a P problem.
 SAT is NP-complete.
 It is the first NP-complete
problem.
 Every NP problem reduces
to SAT.
Stephen Arthur
Cook
(1939~)

21. 8- 21
Transforming searching to SAT
 Does there exist a number in { x(1),
x(2), …, x(n) }, which is equal to 7?
 Assume n = 2.
nondeterministic algorithm:
i = choice(1,2)
if x(i)=7 then SUCCESS
else FAILURE

22. 8- 22
i=1 v i=2
& i=1 → i≠2
& i=2 → i≠1
& x(1)=7 & i=1 → SUCCESS
& x(2)=7 & i=2 → SUCCESS
& x(1)≠7 & i=1 → FAILURE
& x(2)≠7 & i=2 → FAILURE
& FAILURE → -SUCCESS
& SUCCESS (Guarantees a successful
termination)
& x(1)=7 (Input Data)
& x(2)≠7

23. 8- 23
 CNF (conjunctive normal form) :
i=1 v i=2 (1)
i≠1 v i≠2 (2)
x(1)≠7 v i≠1 v SUCCESS (3)
x(2)≠7 v i≠2 v SUCCESS (4)
x(1)=7 v i≠1 v FAILURE (5)
x(2)=7 v i≠2 v FAILURE (6)
-FAILURE v -SUCCESS (7)
SUCCESS (8)
x(1)=7 (9)
x(2)≠7 (10)

24. 8- 24
 Satisfiable at the following assignment :
i=1 satisfying (1)
i≠2 satisfying (2), (4) and (6)
SUCCESS satisfying (3), (4) and (8)
−FAILURE satisfying (7)
x(1)=7 satisfying (5) and (9)
x(2)≠7 satisfying (4) and (10)

25. 8- 25
The semantic tree
i=1 v i=2 (1)
i≠1 v i≠2 (2)
x(1)≠7 v i≠1 v SUCCESS (3)
x(2)≠7 v i≠2 v SUCCESS (4)
x(1)=7 v i≠1 v FAILURE (5)
x(2)=7 v i≠2 v FAILURE (6)
-FAILURE v -SUCCESS (7)
SUCCESS (8)
x(1)=7 (9)
x(2)≠7 (10)

26. 8- 26
Searching for 7, but x(1)≠7, x(2)≠7

CNF (conjunctive normal form) :
i=1 v i=2 (1)
i≠1 v i≠2 (2)
x(1)≠7 v i≠1 v SUCCESS (3)
x(2)≠7 v i≠2 v SUCCESS (4)
x(1)=7 v i≠1 v FAILURE (5)
x(2)=7 v i≠2 v FAILURE (6)
SUCCESS (7)
-SUCCESS v -FAILURE (8)
x(1) ≠ 7 (9)
x(2) ≠ 7 (10)

27. 8- 27
 Apply resolution principle :
(9) & (5) i≠1 v FAILURE (11)
(10) & (6) i≠2 v FAILURE (12)
(7) & (8) -FAILURE (13)
(13) & (11) i≠1 (14)
(13) & (12) i≠2 (15)
(14) & (1) i=2 (11)
(15) & (16) □ (17)
We get an empty clause ⇒ unsatisfiable
⇒ 7 does not exit in x(1) or x(2).

28. 8- 28
 CNF:
i=1 v i=2 (1)
i≠1 v i≠2 (2)
x(1)≠7 v i≠1 v S U C C E S S (3 )
x(2)≠7 v i≠2 v S U C C E S S (4 )
x(1)=7 v i≠1 v FA I L U R E ( 5 )
x(2)=7 v i≠2 v FA I L U R E ( 6 )
SUCCESS (7)
-SUCCESS v -FAILURE (8)
x(1)=7 (9)
x(2)=7 (10)
Searching for 7, where x(1)=7, x(2)=7

29. 8- 29
The semantic
tree
It implies that both assignments (i=1, i=2) satisfy the
clauses.

30. 8- 30
The node cover problem
 Def: Given a graph G=(V, E), S is the node
cover if S ⊆ V and for every edge (u, v) ∈ E,
either u ∈ S or v ∈ S.
node cover :
{1, 3}
{5, 2, 4}
 Decision problem : ∃ S ∋ | S | ≤ K ?

31. 8- 31
Transforming the node cover
problem to SAT
BEGIN
i1 ← choice({1, 2, …, n})
i2 ← choice({1, 2, …, n} – {i1})

ik ← choice({1, 2, …, n} – {i1, i2, …, ik-1}).
For j=1 to m do
BEGIN
if ej is not incident to one of νit
(1≤t≤k)
then FAILURE
END
SUCCESS

32. 8- 32
i1 = 1 v i1 = 2… v i1 = n
(i1≠ 1→i1＝2 v i1＝3…v i1 = n)
i2 = 1 v i2 = 2… v i2 = n

ik = 1 v ik = 2… v ik = n
i1 ≠1 v i2 ≠1 (i1=1 →i2≠1 & ... & ik≠1)
i1 ≠1 v i3 ≠1

ik-1 ≠n v ik ≠n
νi1
∈e1 v νi2
∈e1 v … v νik
∈e1 v FAILURE
(νi1
∉e1&νi2
∉e1&…&νik
∉e1→ Failure)
νi1
∈e2 v νi2
∈e2 v … v νik
∈e2 v FAILURE

νi1
∈em v νi2
∈em v … v νik
∈em v FAILURE
SUCCESS
CN
F:
(To be continued)

33. 8- 33
-SUCCESS v -FAILURE
νr1
∈e1
νs1
∈e1
νr2
∈e2
νs2
∈e2

νrm
∈em
νsm
∈em

34. 8- 34
SAT is NP-complete
(1) SAT has an NP algorithm.
(2) SAT is NP-hard:
 Every NP algorithm for problem A can be
transformed in polynomial time to SAT
[Horowitz 1998] such that SAT is satisfiable
if and only if the answer for A is “YES”.
 That is, every NP problem ∝ SAT .
 By (1) and (2), SAT is NP-complete.

35. 8- 35
Proof of NP-Completeness
 To show that A is NP-complete
(I) Prove that A is an NP problem.
(II) Prove that ∃ B ∈ NPC, B ∝ A.
⇒ A ∈ NPC
 Why ?

36. 8- 36
3-satisfiability problem (3-
SAT)
 Def: Each clause contains exactly three
literals.
 (I) 3-SAT is an NP problem (obviously)
 (II) SAT ∝ 3-SAT
Proof:
(1) One literal L1
in a clause in SAT :
in 3-SAT :
L1
v y1
v y2
L1
v -y1
v y2
L1
v y1
v -y2
L1
v -y1
v -y2

37. 8- 37
(2) Two literals L1
, L2
in a clause in SAT :
in 3-SAT :
L1
v L2
v y1
L1
v L2
v -y1
(3) Three literals in a clause : remain unchanged.
(4) More than 3 literals L1
, L2
, …, Lk
in a clause :
in 3-SAT :
L1
v L2
v y1
L3
v -y1
v y2

Lk-2
v -yk-4
v yk-3
Lk-1
v Lk
v -yk-3

38. 8- 38
SAT transform
3-SAT
S S′
 The instance S′ in 3-
SAT :
x1
v x2
v y1
x1
v x2
v -y1
-x3
v y2
v y3
-x3
v -y2
v y3
-x3
v y2
v -y3
-x3
v -y2
v -y3
x1
v -x2
v y4
x3
v -y4
v y5
-x4
v -y5
v y6
x5
v x6
v -y6
 An instance S in SAT :
x1
v x2
-x3
x1
v -x2
v x3
v -x4
v x5
v x6
Example of transforming SAT to 3-
SAT

39. 8- 39
 Proof : S is satisfiable ⇔ S′ is satisfiable
“⇒”
≤ 3 literals in S (trivial)
consider ≥ 4 literals
S : L1
v L2
v … v Lk
S′: L1
v L2
v y1
L3
v -y1
v y2
L4
v -y2
v y3

Lk-2
v -yk-4
v yk-3
Lk-1
v Lk
v -yk-3

40. 8- 40
 S is satisfiable ⇒ at least Li
= T
Assume : Lj
= F ∀ j ≠ i
assign : yi-1
= F
yj
= T ∀ j < i-1
yj
= F ∀ j > i-1
( Li
v -yi-2
v yi-1
)
⇒ S′ is satisfiable.
 “⇐”
If S′ is satisfiable, then assignment satisfying
S′ can not contain yi
’s only.
⇒ at least one Li
must be true.
(We can also apply the resolution principle).
Thus, 3-SAT is NP-complete.

41. 8- 41
Comment for 3-SAT
 If a problem is NP-complete, its special cases
may or may not be NP-complete.

42. 8- 42
Chromatic number decision
problem (CN)
 Def: A coloring of a graph G=(V, E) is a function
f : V → { 1, 2, 3,…, k } such that if (u, v) ∈ E, then
f(u)≠f(v). The CN problem is to determine if G has a
coloring for k.
<Theorem> Satisfiability with at most 3 literals per
clause (SATY) ∝ CN.
3-colorable
f(a)=1, f(b)=2, f(c)=1
f(d)=2, f(e)=3

43. 8- 43
Proof :
instance of SATY :
variable : x1, x2, …, xn , n ≥ 4
clause : c1, c2, …, cr
instance of CN :
G=(V, E)
V={ x1, x2, …, xn }∪{ -x1, -x2, …, -xn }
∪{ y1, y2, …, yn }∪{ c1, c2, …, cr }

newly added
E={ (xi, -xi) | 1≤ i ≤ n }∪{ (yi, yj) | i ≠ j }
∪{ (yi, xj) | i ≠ j }∪{ (yi, -xj) | i ≠ j }
∪{ (xi, cj) | xi ∉ cj }∪{ (-xi, cj) | -xi ∉ cj }
SATY ∝ CN

44. 8- 44
x1
v x2
v x3
(1)
-x3
v -x4
v x2
(2)
⇓
Example of SATY ∝ CN
True assignment:
x1
=T
x2
=F
x3
=F
x4
=T
E={ (xi
, -xi
) | 1≤ i ≤ n }∪{ (yi
, yj
) | i ≠ j }
∪{ (yi
, xj
) | i ≠ j }∪{ (yi
, -xj
) | i ≠ j }
∪{ (xi
, cj
) | xi
∉ cj
}∪{ (-xi
, cj
) | -xi
∉ cj
}

45. 8- 45
 Satisfiable ⇔ n+1 colorable
 “⇒”
(1) f(yi
) = i
(2) if xi
= T, then f(xi
) = i, f(-xi
) = n+1
else f(xi
) = n+1, f(-xi
) = i
(3)if xi
in cj
and xi
= T, then f(cj
) = f(xi
)
if -xi
in cj
and -xi
= T, then f(cj
) = f(-xi
)
( at least one such xi
)
Proof of SATY ∝ CN

46. 8- 46
 “⇐”
(1) yi
must be assigned with color i.
(2) f(xi
) ≠ f(-xi
)
either f(xi
) = i and f(-xi
) = n+1
or f(xi
) = n+1 and f(-xi
) = i
(3) at most 3 literals in cj
and n ≥ 4
⇒ at least one xi
, ∋ xi
and -xi
are not in cj
⇒ f(cj
) ≠ n+1
(4) if f(cj
) = i = f(xi
), assign xi
to T
if f(cj
) = i = f(-xi
), assign -xi
to T
(5) if f(cj
) = i = f(xi
) ⇒ (cj
, xi
) ∉ E
⇒ xi
in cj
⇒ cj
is true
if f(cj
) = i = f(-xi
) ⇒ similarly

47. 8- 47
Set cover decision problem
 Def: F = {Si
} = { S1
, S2
, …, Sk
}
= { u1
, u2
, …, un
}
T is a set cover of F if T ⊆ F and
The set cover decision problem is to determine if F has
a cover T containing no more than c sets.
 Example
F = {(u1
, u3
), (u2
, u4
), (u2
, u3
), (u4
), (u1
, u3
, u4
)}
s1
s2
s3
s4
s5
T = { s1
, s3
, s4
} set cover
T = { s1
, s2
} set cover, exact cover
FS
i
i
S
∈

FS
i
TS
i
ii
SS
∈∈
=

48. 8- 48
Exact cover problem
(Notations same as those in set cover.)
Def: To determine if F has an exact cover T,
which is a cover of F and the sets in T are
pairwise disjoint.
<Theorem> CN ∝ exact cover
(No proof here.)

49. 8- 49
Sum of subsets problem
 Def: A set of positive numbers A = { a1
, a2
, …,
an
}
a constant C
Determine if ∃ A′ ⊆ A ∋
 e.g. A = { 7, 5, 19, 1, 12, 8, 14 }
 C = 21, A′ = { 7, 14 }
 C = 11, no solution
<Theorem> Exact cover ∝ sum of subsets.
Ca
Aa
i
i
=∑′∈

50. 8- 50
 Proof :
instance of exact cover :
F = { S1
, S2
, …, Sk
}
instance of sum of subsets :
A = { a1
, a2
, …, ak
} where
where eij
= 1 if uj
∈ Si
eij
= 0 if otherwise.
 Why k+1? (See the example on the next page.)
Exact cover ∝ sum of subsets
{ }
FS
ni
i
uuuS
∈
= ...,,2,1
j
nj
iji kea )1(
1
+= ∑≤≤
kkkkC n
nj
j
/)1)1)((1()1(
1
−++=+= ∑≤≤

51. 8- 51
Example of Exact cover ∝ sum of
subsets
 Valid transformation:
u1=6, u2=8, u3=9, n=3
EC: S1={6,8}, S2={9},
S3={6,9}, S4={8,9}
k=4
SS: a1=51
+52
=30
a2=53
=125
a3=51
+53
=130
a4=52
+53
=150
 Invalid transformation:
EC: S1={6,8}, S2={8}, S3={8},
S4={8,9}. K=4
Suppose k-2=2 is used.
SS: a1=21
+22
=6
a2=22
=4
a3=22
=4
a4=22
+23
=12
C=21
+22
+23
=14
{ } { }9,8,6...,,2,1 ==
∈

FS
ni
i
uuuS

52. 8- 52
Partition problem
 Def: Given a set of positive numbers A =
{ a1
,a2
,…,an
},
determine if ∃ a partition P, ∋
 e. g. A = {3, 6, 1, 9, 4, 11}
partition : {3, 1, 9, 4} and {6, 11}
<Theorem> sum of subsets ∝ partition
∑∑ ∉∈
=
Pa
i
Pa
i
ii
aa

53. 8- 53
Sum of subsets ∝ partition
proof :
instance of sum of subsets :
A = { a1, a2, …, an }, C
instance of partition :
B = { b1, b2, …, bn+2 }, where bi = ai, 1≤ i ≤ n
bn+1 = C+1
bn+2 = ( ∑ ai )+1−C
1≤i≤n
C = ∑ai ⇔ ( ∑ai )+bn+2 = ( ∑ai )+bn+1
ai∈S ai∈S ai∉S
⇔ partition : { bi | ai∈S}∪{bn+2}
and { bi | ai∉S }∪{bn+1}
S
S’
A C

54. 8- 54
 Why bn+1
= C+1 ? why not bn+1
= C ?
 To avoid bn+1
and bn+2
to be partitioned into
the same subset.

55. 8- 55
Bin packing problem
 Def: n items, each of size ci
, ci
> 0
Each bin capacity : C
 Determine if we can assign the items into
k bins, ∋ ∑ci
≤ C , 1≤j≤k.
i∈binj
<Theorem> partition ∝ bin packing.

56. 8- 56
VLSI discrete layout problem
 Given: n rectangles, each with height hi
(integer)
width wi
and an area A
Determine if there is a placement of the n
rectangles within the area A according to the rules :
1. Boundaries of rectangles parallel to x axis or y
axis.
2. Corners of rectangles lie on integer points.
3. No two rectangles overlap.
4. Two rectangles are separated by at least a unit
distance.
(See the figure on the next page.)

57. 8- 57
A Successful Placement
<Theorem> bin packing ∝ VLSI discrete layout.

58. 8- 58
Max clique problem
 Def: A maximal complete subgraph of a graph
G=(V,E) is a clique. The max (maximum) clique
problem is to determine the size of a largest
clique in G.
 e. g.
<Theorem> SAT ∝ clique decision problem.
maximal cliques :
{a, b}, {a, c, d}
{c, d, e, f}
maximum clique :
(largest)
{c, d, e, f}

59. 8- 59
Node cover decision problem
 Def: A set S ⊆ V is a node cover for a graph
G = (V, E) iff all edges in E are incident to at
least one vertex in S. ∃ S, ∋ |S| ≤ K ?
<Theorem> clique decision problem ∝
node cover decision problem.
(See proof on the next page.)

60. 8- 60
Clique decision ∝ node cover
decision
 G=(V,E) : clique Q of size k (Q⊆V)
G’=(V,E’) : node cover S of size n-k, S=V-Q
where E’={(u,v)|u∈V, v ∈V and (u,v)∉E}

61. 8- 61
Hamiltonian cycle problem
 Def: A Hamiltonian cycle is a round trip path
along n edges of G which visits every vertex
once and returns to its starting vertex.
 e.g.
Hamiltonian cycle : 1, 2, 8, 7, 6, 5, 4, 3, 1.
<Theorem> SAT ∝ directed Hamiltonian cycle
( in a directed graph )

62. 8- 62
Traveling salesperson problem
 Def: A tour of a directed graph G=(V,
E) is a directed cycle that includes every
vertex in V. The problem is to find a tour
of minimum cost.
<Theorem> Directed Hamiltonian cycle ∝
traveling salesperson decision problem.
(See proof on the next page.)

63. 8- 63
Proof of Hamiltonian ∝ TSP

64. 8- 64
0/1 knapsack problem
 Def: n objects, each with a weight wi
> 0
a profit pi
> 0
capacity of knapsack : M
Maximize ∑pi
xi
1≤i≤n
Subject to ∑wi
xi
≤ M
1≤i≤n
xi
= 0 or 1, 1≤ i ≤n
 Decision version :
Given K, ∃ ∑pi
xi
≥ K ?
1≤i≤n
 Knapsack problem : 0 ≤ xi
≤ 1, 1≤ i ≤n.
<Theorem> partition ∝ 0/1 knapsack decision
problem.

65. 8- 65
 Refer to Sec. 11.3, Sec. 11.4 and its
exercises of [Horowitz 1998] for the proofs of
more NP-complete problems.
 [[Horowitz 1998] E. Howowitz, S. Sahni and S.
Rajasekaran, Computer Algorithms, Computer
Science Press, New York, 1998, 「台北圖書」代
理 , 02-23625376